An urn contains 4 black and 6 white balls. Two balls are drawn
without replacement. All black balls in this sample are reserved. All white
balls (if any) are returned to the urn and the same number of balls is drawn
from the urn. What is the total expected number of black balls obtained in
this experiment?
So I was thinking in terms of total expectation where X is the total number of black balls and Y is the number of white balls in the first draw:
$E(X) = E(E(X|Y)) = (4/10)(3/9) E(X|Y=0) + (6/10)(4/9)*2*E(X|Y=1) + (6/10)(5/10)E(X|Y=2) $
Is this the right track? Because indicator variables seem to be possible here too.
Best Answer
$E\left(X\right)=E\left(X\mid Y=0\right)P\left\{ Y=0\right\} +E\left(X\mid Y=1\right)P\left\{ Y=1\right\} +E\left(X\mid Y=2\right)P\left\{ Y=2\right\} $
Here $P\left\{ Y=0\right\} =\binom{10}{2}^{-1}\binom{4}{2}\binom{6}{0}=\frac{2}{15}$ , $P\left\{ Y=1\right\} =\binom{10}{2}^{-1}\binom{4}{1}\binom{6}{1}=\frac{8}{15}$ and $P\left\{ Y=2\right\} =\binom{10}{2}^{-1}\binom{4}{0}\binom{6}{2}=\frac{5}{15}$
This agrees with what you are saying in your question (with exception of the typo mentioned by André).
$E\left(X\mid Y=0\right)=2+EU$ where $U$ is the number of obtained blacks by drawing $2$ balls from an urn with $0$ blacks and $6$ whites.
$E\left(X\mid Y=1\right)=1+EV$ where $V$ is the number of obtained blacks by drawing $2$ balls from an urn with $1$ blacks and $6$ whites.
$E\left(X\mid Y=2\right)=0+EW$ where $W$ is the number of obtained blacks by drawing $2$ balls from an urn with $2$ blacks and $6$ whites.
It is clear that $U=0$, hence $EU=0$. As André points out in his comment you could also agree that in this situation the second drawing does not take place. There are no black balls to catch anymore. Next to that $EV=\frac{2}{7}$ and $EW=\frac{1}{2}$. They can be calculated as mean of hypergeometric distributions (or more directly by using indicator rv's). This leads to:
$EX=\left(2+0\right)\frac{2}{15}+\left(1+\frac{2}{7}\right)\frac{8}{15}+\left(0+\frac{1}{2}\right)\frac{5}{15}=\frac{47}{42}$